a definition : tomic polynomial

[tommy1729]

just wanted to explain my definition "tomic polynomial"

it might already have a name ; i am unaware of that ...

the reason i post this is because i like algebra and perhaps someone can learn me something nice about these "tomic polynomials". Or someone can give me the real name of them , so that i can look for info on them on the internet e.g.

( ternary polynomial maybe ? )

its also plausable that this concept might arise in the future in one of my threaths , so ill clarify in advance.

i hope it is clearly stated by me...

***
tomic polynomial:

all coefficients are E [-1,0,1]

none of its zero's is a root of unity / [-1,1]

zero is not a root of the polynomial
***

regards
tommy1729

[quasi]

On Thu, 23 Aug 2007 17:14:46 EDT, tommy1729 <tomm...@gmail.com>
wrote:

Conjecture:

Every reducible tomic polynomial has at least one nonconstant
irreducible tomic factor.

quasi

[Brian VanPelt]

What's the ring here? Are -1, 0, and 1 considered to be in the
complex number system, or are they the multiplicative identity (and
its additive inverse) in any ring - along with the additive identity?

Irrecducibility varies depending on the ring structure.

Now, if your coefficients are considered as complex numbers, then the
Fundamental Theorem of Algebra affirms your conjecture.

I haven't seen the post prior to this, so I might be missing something
simple.

Thanks,

Brian

[quasi]

Yes, true.

In this context, the intended assumption was the ring of polynomials
with integer coefficients. But I should have made that explicit.

>Now, if your coefficients are considered as complex numbers, then the
>Fundamental Theorem of Algebra affirms your conjecture.

For tomic polynomials, the nonzero coefficients are all 1 or -1, as
declared in tommy's definition (as quoted above).

As far as I can see, the FTA does yield anything obvious with respect
to proving or disproving the conjecture.

>I haven't seen the post prior to this, so I might be missing something
>simple.

Yes, you misunderstood the definition.

quasi

[quasi]

On Fri, 24 Aug 2007 22:31:24 -0400, quasi <qu...@null.set> wrote:

Well, Maple kills my conjecture ...

The tomic polynomial

x^12 + x^11 + x^9 - x^8 + x^6 - x^4 - x^3 - x + 1

factors into two irreducible polynomials

(x^6 - x^5 + x^4 - x^2 + 2x - 1) (x^6 + 2x^5 + x^4 - x^2 - x - 1)

neither of which is tomic.

quasi

[tommy1729]

naive conjecture 1:

( all polynomials univariate of course )

conditions:

let p_1 be an integer polynomial ( integer coefficients ) and let p_1 be a non-tomic polynomial.

also the coefficients of p_1 have no gcd > 1 and p_1 has no roots of unity nor zero.

p_2 is an integer polynomial ( integer coefficients ) , has no roots of unity nor zero.

p_1 has as many non-zero coefficients as its degree , and its degree is >= 5.

statement : for any given p_1 there exists a p_2 such that :

p_1(x) * p_2(x) = tomic polynomial(x)

or p_1(x) / p_2(x) = tomic polynomial(x)

***

perhaps too naive and impulsive , but im curious what answers , proofs , disproofs and modifications will be made by others ....

regards
tommy1729

[tommy1729]

naive conjecture 2:

there is a simple pattern in the list of tomic polynomials ordered by degree and coefficients order -1,0,1

and q(n)= the amount of tomic polynomials of degree n or below can be done in closed form.

regards
tommy1729

[tommy1729]

(x -/+ 1)^2 * poly(x) =/= tomic

tommy1729

[amy666]

Every algebraic number is a ( finite ) sum of zero's of tommy polynomials.

regards

tommy1729